Fzu 2171 line segment tree Interval Update

Http://acm.fzu.edu.cn/problem.php? PID = 1, 2171

 

Problem 2171 defensive position IIAccept: 105 submit: 415
Time Limit: 3000 msec memory limit: 32768 kb Problem Description

There are a total of N soldiers in the army, each of which has its own capacity index XI. During a drill, the headquarters determined m locations to be defended, the headquarters will select M soldiers to go to the designated location for defensive tasks in turn. The reference index obtained is the sum of the capabilities of M soldiers. Over time, the headquarters will issue Q commands to replace m of soldiers defending. The performance index of each soldier who completes the defensive task will drop by 1 due to fatigue and other reasons. Now the soldiers are in a row. Please calculate the reference index of each defender.

Input

The input contains multiple groups of data.

The first line contains two integers, n, m, and Q (1 <=n <= 100000,1 <= m <= 100000 ,1 <= q <= ), the N integers in the second row indicate the capability index XI (1 <= xi <= 1000) of each soldier ).

In the next Q row, each row has an integer x, indicating that the previous soldiers are replaced with M soldiers starting with X in the original queue. (1 <= x <= N-M + 1)

For 30% of data 1 <= m, n, q <= 1000.

Output

Output Q rows, each row has an integer, which is the reference index of soldiers who defend each command execution.

Sample input5 3 32 1 3 1 4123 sample output635 ============================ ========================================== classic line segment tree interval updates, only one code is posted here. You still need to understand it.

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#define Maxx 100010using namespace std;int str[Maxx];long long ans;struct Node{    int left,right,add,maxnode,sum;};Node tree[Maxx<<2];void build(int l,int r,int rt){    tree[rt].left=l;    tree[rt].right=r;    tree[rt].add=0;    if(l == r)    {        tree[rt].sum=str[l];        return ;    }    int mid=(l+r)/2;    build(l,mid,rt*2);    build(mid+1,r,rt*2+1);    //return 0;    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}void update(int l,int r,int rt,int add){    if(tree[rt].left>r || tree[rt].right<l)    {        return ;    }    if(tree[rt].left >=l&&tree[rt].right <=r)    {        tree[rt].sum+=(tree[rt].right-tree[rt].left+1)*add;        tree[rt].add+=add;        return ;    }    if(tree[rt].add)    {        tree[rt*2].sum+=(tree[rt*2].right-tree[rt*2].left+1)*tree[rt].add;        tree[rt*2].add+=tree[rt].add;        tree[rt*2+1].sum+=(tree[rt*2+1].right-tree[rt*2+1].left+1)*tree[rt].add;        tree[rt*2+1].add+=tree[rt].add;        tree[rt].add=0;    }    update(l,r,rt*2,add);    update(l,r,rt*2+1,add);    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}void query(int l,int r,int rt){    if(tree[rt].left>r || tree[rt].right<l)    {        return ;    }    if(tree[rt].left>=l && tree[rt].right<=r)    {        ans+=tree[rt].sum;        return ;    }    if(tree[rt].add)    {        tree[rt*2].sum+=(tree[rt*2].right-tree[rt*2].left+1)*tree[rt].add;        tree[rt*2].add+=tree[rt].add;        tree[rt*2+1].sum+=(tree[rt*2+1].right-tree[rt*2+1].left+1)*tree[rt].add;        tree[rt*2+1].add+=tree[rt].add;        tree[rt].add=0;    }    query(l,r,rt*2);    query(l,r,rt*2+1);    tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;    //return ans;}int main(){    int n,m,q;    int x,i,j;    while(scanf("%d%d%d",&n,&m,&q)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&str[i]);        }        build(1,n,1);        for(i=0;i<q;i++)        {            scanf("%d",&x);            ans=0;            query(x,x+m-1,1);            printf("%lld\n",ans);            update(x,x+m-1,1,-1);        }    }    return 0;}